Kabir wants to know the volume of a solid right pyramid

Find perimeter pt4 brainliest

Irina18 [472]

Perimeter= X + Y + W. Perimeter is the sum of all sides

8 0

2 years ago

Convert 0.50 sen /g to RM/kg

IgorC [24]

the conversion is:

0.50 sen /g = 5 RM/kg

<h3>How to change the units?</h3>

First, we know that

1sen = 0.01RM

Then we can change:

0.50 sen /g = 0.50*(0.01 RM) /g = 0.005 RM/g

Now we use the relation:

1g = 0.001kg

So we can rewrite:

0.005 RM/g = 0.005 RM/(0.001kg) = 5 RM/kg

So the conversion is:

0.50 sen /g = 5 RM/kg

If you want to learn more about changes of units:

brainly.com/question/9032119

#SPJ1

7 0

2 years ago

Suppose you need to get the dent in your car fixed. You see the ads in the newspaper above (mikes repair shop $100 service charg

Norma-Jean [14]

Answer:

Total Cost at Amy's shop is $280 which is less than the total cost at Mike's shop which is $295, so I will take the car to Amy's shop

Point of intersection: (x,y) = (4,360)

Explanation:

Part 1 )

Let x = number of hours worked

Let y = total cost to you

Determining where to take the car:

Total Cost at Amy's shop is $280 which is less than the total cost at Mike's shop which is $295

Mike's repair shop:

Service fee: $100

Per hour rate: $65

Let x is the number of hours worked on. Total cost will be equal to the service fee plus hourly charges. Charges per an hour are $65, so for x hours the charges will be 65x.

Therefore, total cost for Amy's repair shop can be written as:

y = 100 + 65x ------(1)

For 3 hours work: y = 100 + 65(3)

Cost at Mike's, y = $100 + $195 = $295

Amy's repair shop: 

Service fee: $40

Per hour rate: $80

Charges per an hour are $40, so for x hours the charges will be 80x.

Therefore, total cost for Amy's repair shop can be written as:

y = 40 + 80x ------(2)

For 3 hours work: x = 3

y = 40 + 80(3)

Cost at Amy's, y = $40 + $240 = $280

Part 2)

Find the point of intersection

Using elimination method:

y = 100 + 65 x ------1

y = 40 + 80x ------2

As the y coefficients are equal we will subtract eq1 from eq2

y - y = 40 + 80x - 100 - 65x

0 = -60 + 15x

60 = 15x

x = $\frac{60}{4}$

x = 4

put x = 4 in eq 1

we get y = 100 + 65 (4) = 100 + 260 = 360

x = 4, y = 360

Point of intersection: (x,y) = (4,360)

8 0

3 years ago

George places 35 books on 7 shelves each shelf holds the same number of books which equations can be used to find how many books

tino4ka555 [31]

35 divided by 7=5 so there's 5 books on each shelf

5 0

3 years ago

Choose 5 cards from a full deck of 52 cards with 13values (2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A) and 4 kinds(spade, diamond, h

Delvig [45]

Answer:

a) 182 possible ways.

b) 5148 possible ways.

c) 1378 possible ways.

d) 2899 possible ways.

Step-by-step explanation:

The order in which the cards are chosen is not important, which means that we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question, we have that:

There are 52 total cards, of which:

13 are spades.

13 are diamonds.

13 are hearts.

13 are clubs.

(a)Two-pairs: Two pairs plus another card of a different value, for example:

2 pairs of 2 from sets os 13.

1 other card, from a set of 26(whichever two cards were not chosen above). So

$T = 2C_{13,2} + C_{26,1} = 2*\frac{13!}{2!11!} + \frac{26!}{1!25!} = 182$

So 182 possible ways.

(b)Flush: five cards of the same suit but different values, for example:

4 combinations of 5 from a set of 13(can be all spades, all diamonds, and hearts or all clubs). So

$T = 4*C_{13,5} = 4*\frac{13!}{5!8!} = 5148$

So 5148 possible ways.

(c)Full house: A three of a kind and a pair, for example:

4 combinations of 3 from a set of 13(three of a kind ,c an be all possible kinds).

3 combinations of 2 from a set of 13(the pair, cant be the kind chosen for the trio, so 3 combinations). So

$T = 4*C_{13,3} + 3*C_{13.2} = 4*\frac{13!}{3!10!} + 3*\frac{13!}{2!11!} = 1378$

So 1378 possible ways.

(d)Four of a kind: Four cards of the same value, for example:

4 combinations of 4 from a set of 13(four of a kind, can be all spades, all diamonds, and hearts or all clubs).

1 from the remaining 39(do not involve the kind chosen above). So

$T = 4*C_{13,4} + C_{39,1} = 4*\frac{13!}{4!9!} + \frac{39!}{1!38!} = 2899$

So 2899 possible ways.

4 0

3 years ago